[proofplan] The proof reduces the stability of the real system to growth bounds for the matrix exponential $e^{tA}$. After complexifying $A$, we put the complex matrix into [Jordan normal form](/theorems/864) and compute the exponential of each Jordan block explicitly. The exponential factor $e^{t\operatorname{Re}(\lambda)}$ controls decay or growth, while nontrivial nilpotent parts contribute polynomial factors. Combining the block estimates through the change-of-basis matrix gives the three stability classifications. [/proofplan] [step:Express stability in terms of the matrix exponential] Define the matrix exponential map \begin{align*} E_A: [0,\infty) \to \mathbb{R}^{n \times n}, \qquad t \mapsto e^{tA} := \sum_{k=0}^{\infty} \frac{t^k A^k}{k!}. \end{align*} The series converges absolutely in every matrix norm for each $t \geq 0$. Termwise differentiation is justified by [uniform convergence](/page/Uniform%20Convergence) on compact time intervals, and gives \begin{align*} \frac{d}{dt}e^{tA} = A e^{tA}. \end{align*} Also $e^{0A}=I_n$. Hence, for each initial value $x_0 \in \mathbb{R}^n$, the map \begin{align*} x_{x_0}: [0,\infty) \to \mathbb{R}^n, \qquad t \mapsto e^{tA}x_0 \end{align*} solves $x'(t)=Ax(t)$ and $x(0)=x_0$. We use the Euclidean norm $|\cdot|$ on $\mathbb{R}^n$ and the induced operator norm $\|\cdot\|_{\mathrm{op}}$ on $\mathbb{R}^{n \times n}$. The equilibrium $0$ is exponentially stable exactly when there exist constants $M \geq 1$ and $\alpha > 0$ such that \begin{align*} |e^{tA}x_0| \leq M e^{-\alpha t}|x_0| \end{align*} for all $t \geq 0$ and all $x_0 \in \mathbb{R}^n$. Equivalently, \begin{align*} \|e^{tA}\|_{\mathrm{op}} \leq M e^{-\alpha t} \end{align*} for all $t \geq 0$. The equilibrium $0$ is Lyapunov stable exactly when there exists a constant $K \geq 1$ such that \begin{align*} \|e^{tA}\|_{\mathrm{op}} \leq K \end{align*} for all $t \geq 0$. Indeed, such a bound gives $|e^{tA}x_0| \leq K|x_0|$ for every $t \geq 0$. Conversely, Lyapunov stability with $\varepsilon = 1$ gives a number $\delta > 0$ such that $|x_0| < \delta$ implies $|e^{tA}x_0| < 1$ for all $t \geq 0$. By homogeneity, for every nonzero $x_0 \in \mathbb{R}^n$, \begin{align*} |e^{tA}x_0| = \frac{2|x_0|}{\delta}\left|e^{tA}\left(\frac{\delta x_0}{2|x_0|}\right)\right| \leq \frac{2}{\delta}|x_0|. \end{align*} Thus $K := \max\{1,2/\delta\}$ works. Finally, asymptotic stability is equivalent to Lyapunov stability together with \begin{align*} \lim_{t \to \infty} e^{tA}x_0 = 0 \end{align*} for every $x_0 \in \mathbb{R}^n$. [/step] [step:Compute the exponential on each complex Jordan block] Let $A_{\mathbb{C}} \in \mathbb{C}^{n \times n}$ denote the complexification of $A$, acting on $\mathbb{C}^n$. We use the Hermitian Euclidean norm $|\cdot|_{\mathbb{C}^n}$ and the induced operator norm $\|\cdot\|_{\mathcal{L}(\mathbb{C}^n)}$. Complexification does not change the eigenvalues of $A$, and boundedness or exponential decay of $e^{tA}$ on $\mathbb{R}^n$ is equivalent to the corresponding property of $e^{tA_{\mathbb{C}}}$ on $\mathbb{C}^n$, up to the same finite-dimensional norm comparison. By the complex [Jordan normal form theorem](/theorems/412) (citing a result not yet in the wiki: complex Jordan normal form), there exist an invertible matrix $S \in GL(n,\mathbb{C})$, complex numbers $\lambda_1,\dots,\lambda_r \in \mathbb{C}$, positive integers $m_1,\dots,m_r \in \mathbb{N}$ with $m_1+\cdots+m_r=n$, and nilpotent Jordan matrices $N_j \in \mathbb{C}^{m_j \times m_j}$ satisfying $N_j^{m_j}=0$, such that \begin{align*} A_{\mathbb{C}} = S J S^{-1}, \end{align*} where $J$ is block diagonal with blocks \begin{align*} J_j = \lambda_j I_{m_j} + N_j. \end{align*} Here $I_{m_j}$ denotes the $m_j \times m_j$ identity matrix. Since the matrix exponential is defined by a [power series](/page/Power%20Series), it respects similarity and block diagonal decompositions: \begin{align*} e^{tA_{\mathbb{C}}} = S e^{tJ} S^{-1}. \end{align*} For each block $J_j$, the matrices $\lambda_j I_{m_j}$ and $N_j$ commute, so \begin{align*} e^{tJ_j} = e^{t\lambda_j} e^{tN_j}. \end{align*} Because $N_j^{m_j}=0$, the nilpotent exponential is the finite sum \begin{align*} e^{tN_j} = \sum_{k=0}^{m_j-1}\frac{t^k N_j^k}{k!}. \end{align*} Therefore \begin{align*} e^{tJ_j} = e^{t\lambda_j}\sum_{k=0}^{m_j-1}\frac{t^k N_j^k}{k!}. \end{align*} [guided] The purpose of complexification is to make the block structure completely visible. A real matrix can have complex conjugate eigenvalues, but after passing from $\mathbb{R}^n$ to $\mathbb{C}^n$, every eigenvalue is represented by a Jordan block. Let $A_{\mathbb{C}} \in \mathbb{C}^{n \times n}$ be the same matrix as $A$, now acting complex-linearly on $\mathbb{C}^n$. We invoke the complex Jordan normal form theorem (citing a result not yet in the wiki: complex Jordan normal form). It gives an invertible matrix $S \in GL(n,\mathbb{C})$ and a block diagonal matrix $J$ such that \begin{align*} A_{\mathbb{C}} = S J S^{-1}. \end{align*} Each block has the form \begin{align*} J_j = \lambda_j I_{m_j} + N_j, \end{align*} where $\lambda_j \in \mathbb{C}$ is an eigenvalue, $I_{m_j}$ is the identity matrix on $\mathbb{C}^{m_j}$, and $N_j \in \mathbb{C}^{m_j \times m_j}$ is nilpotent with $N_j^{m_j}=0$. Why is this useful for stability? Because the exponential of a Jordan block separates into an exponential part and a polynomial part. Since the matrix exponential is defined by the absolutely convergent power series \begin{align*} e^{tB} := \sum_{k=0}^{\infty}\frac{t^kB^k}{k!} \end{align*} for a complex square matrix $B$, similarity is preserved: \begin{align*} e^{tA_{\mathbb{C}}} = e^{tSJS^{-1}} = S e^{tJ} S^{-1}. \end{align*} The block diagonal structure of $J$ means $e^{tJ}$ is block diagonal with blocks $e^{tJ_j}$. Now fix a block $J_j=\lambda_j I_{m_j}+N_j$. The scalar matrix $\lambda_j I_{m_j}$ commutes with $N_j$, so the exponential of the sum factors: \begin{align*} e^{tJ_j} = e^{t\lambda_j I_{m_j}}e^{tN_j} = e^{t\lambda_j}e^{tN_j}. \end{align*} The nilpotent part has a finite exponential series because $N_j^{m_j}=0$: \begin{align*} e^{tN_j} = \sum_{k=0}^{m_j-1}\frac{t^kN_j^k}{k!}. \end{align*} Thus \begin{align*} e^{tJ_j} = e^{t\lambda_j}\sum_{k=0}^{m_j-1}\frac{t^kN_j^k}{k!}. \end{align*} This formula is the entire stability mechanism: $e^{t\operatorname{Re}(\lambda_j)}$ decides exponential decay or growth, while the nilpotent terms can add polynomial growth when the Jordan block has size larger than $1$. [/guided] [/step] [step:Show Hurwitz matrices give exponential stability] Assume $A$ is Hurwitz. Define the spectral gap \begin{align*} \beta := -\max_{1 \leq j \leq r}\operatorname{Re}(\lambda_j). \end{align*} Then $\beta>0$. Choose \begin{align*} \alpha := \frac{\beta}{2}. \end{align*} For each block $J_j$, define the polynomial bound \begin{align*} P_j(t) := \sum_{k=0}^{m_j-1}\frac{t^k\|N_j^k\|_{\mathcal{L}(\mathbb{C}^{m_j})}}{k!}, \qquad t \geq 0. \end{align*} Then \begin{align*} \|e^{tJ_j}\|_{\mathcal{L}(\mathbb{C}^{m_j})} \leq e^{t\operatorname{Re}(\lambda_j)}P_j(t) \leq e^{-\beta t}P_j(t). \end{align*} Since $P_j$ is a polynomial and $\beta-\alpha=\beta/2>0$, the constant \begin{align*} C_j := \sup_{t \geq 0} e^{-(\beta-\alpha)t}P_j(t) \end{align*} is finite. Hence \begin{align*} \|e^{tJ_j}\|_{\mathcal{L}(\mathbb{C}^{m_j})} \leq C_j e^{-\alpha t} \end{align*} for every $t \geq 0$. Define \begin{align*} C_J := \max_{1 \leq j \leq r} C_j. \end{align*} Since $e^{tJ}$ is block diagonal, its operator norm is bounded by a finite-dimensional block norm constant times $C_J e^{-\alpha t}$. Absorbing that block norm constant into $C_J$, we have \begin{align*} \|e^{tJ}\|_{\mathcal{L}(\mathbb{C}^n)} \leq C_J e^{-\alpha t}. \end{align*} Therefore \begin{align*} \|e^{tA_{\mathbb{C}}}\|_{\mathcal{L}(\mathbb{C}^n)} \leq \|S\|_{\mathcal{L}(\mathbb{C}^n)}\|S^{-1}\|_{\mathcal{L}(\mathbb{C}^n)}C_J e^{-\alpha t}. \end{align*} Restricting to real vectors gives constants $M \geq 1$ and $\alpha>0$ such that \begin{align*} |e^{tA}x_0| \leq M e^{-\alpha t}|x_0| \end{align*} for all $x_0 \in \mathbb{R}^n$ and $t \geq 0$. Thus $0$ is exponentially stable. [/step] [step:Show exponential stability forces all eigenvalues into the open left half-plane] Assume $0$ is exponentially stable. Then there exist constants $M \geq 1$ and $\alpha>0$ such that \begin{align*} \|e^{tA_{\mathbb{C}}}\|_{\mathcal{L}(\mathbb{C}^n)} \leq M e^{-\alpha t} \end{align*} for all $t \geq 0$ after complexification. Let $\lambda \in \mathbb{C}$ be an eigenvalue of $A_{\mathbb{C}}$, and let $v \in \mathbb{C}^n \setminus \{0\}$ be an eigenvector with $A_{\mathbb{C}}v=\lambda v$. Then \begin{align*} e^{tA_{\mathbb{C}}}v = e^{t\lambda}v. \end{align*} Taking norms gives \begin{align*} e^{t\operatorname{Re}(\lambda)}|v|_{\mathbb{C}^n} \leq M e^{-\alpha t}|v|_{\mathbb{C}^n}. \end{align*} Since $v \neq 0$, division by $|v|_{\mathbb{C}^n}$ yields \begin{align*} e^{t(\operatorname{Re}(\lambda)+\alpha)} \leq M \end{align*} for every $t \geq 0$. This is possible only if $\operatorname{Re}(\lambda)+\alpha \leq 0$. Hence $\operatorname{Re}(\lambda)<0$, and $A$ is Hurwitz. [/step] [step:Derive asymptotic stability from the Hurwitz condition and conversely] If $A$ is Hurwitz, the previous exponential estimate gives \begin{align*} |e^{tA}x_0| \leq M e^{-\alpha t}|x_0| \end{align*} for all $x_0 \in \mathbb{R}^n$ and $t \geq 0$. This implies Lyapunov stability and \begin{align*} \lim_{t \to \infty}e^{tA}x_0 = 0 \end{align*} for every $x_0 \in \mathbb{R}^n$. Thus $0$ is asymptotically stable. Conversely, assume $0$ is asymptotically stable. Suppose, toward a contradiction, that $A$ has an eigenvalue $\lambda \in \mathbb{C}$ with $\operatorname{Re}(\lambda)\geq 0$. Let $v \in \mathbb{C}^n\setminus\{0\}$ be an eigenvector for $\lambda$. Then \begin{align*} e^{tA_{\mathbb{C}}}v = e^{t\lambda}v. \end{align*} If $\operatorname{Re}(\lambda)>0$, then $|e^{tA_{\mathbb{C}}}v|_{\mathbb{C}^n}=e^{t\operatorname{Re}(\lambda)}|v|_{\mathbb{C}^n}$ is unbounded. If $\operatorname{Re}(\lambda)=0$, then $|e^{tA_{\mathbb{C}}}v|_{\mathbb{C}^n}=|v|_{\mathbb{C}^n}$ for all $t \geq 0$, so the orbit does not converge to $0$. Writing $v=a+ib$ with $a,b \in \mathbb{R}^n$, at least one of $a$ and $b$ is nonzero. Since \begin{align*} e^{tA_{\mathbb{C}}}v = e^{tA}a + i e^{tA}b, \end{align*} the convergence of all real solutions to $0$ would imply convergence of $e^{tA_{\mathbb{C}}}v$ to $0$, a contradiction. Hence every eigenvalue satisfies $\operatorname{Re}(\lambda)<0$, so $A$ is Hurwitz. [/step] [step:Classify Lyapunov stability by bounded Jordan blocks] Assume first that every eigenvalue $\lambda$ satisfies $\operatorname{Re}(\lambda)\leq 0$, and every eigenvalue with $\operatorname{Re}(\lambda)=0$ is semisimple. For a Jordan block $J_j=\lambda_j I_{m_j}+N_j$, there are two cases. If $\operatorname{Re}(\lambda_j)<0$, then the same polynomial-versus-exponential estimate used above gives a constant $B_j>0$ such that \begin{align*} \|e^{tJ_j}\|_{\mathcal{L}(\mathbb{C}^{m_j})} \leq B_j \end{align*} for every $t \geq 0$. If $\operatorname{Re}(\lambda_j)=0$, semisimplicity means $m_j=1$ and $N_j=0$ on that block. Hence \begin{align*} e^{tJ_j}=e^{t\lambda_j}I_1, \end{align*} and \begin{align*} \|e^{tJ_j}\|_{\mathcal{L}(\mathbb{C})}=|e^{t\lambda_j}|=e^{t\operatorname{Re}(\lambda_j)}=1 \end{align*} for every $t \geq 0$. Since there are finitely many blocks, there is a constant $B_J>0$ such that \begin{align*} \|e^{tJ}\|_{\mathcal{L}(\mathbb{C}^n)} \leq B_J \end{align*} for every $t \geq 0$. Therefore \begin{align*} \|e^{tA_{\mathbb{C}}}\|_{\mathcal{L}(\mathbb{C}^n)} \leq \|S\|_{\mathcal{L}(\mathbb{C}^n)}\|S^{-1}\|_{\mathcal{L}(\mathbb{C}^n)}B_J. \end{align*} Restricting to $\mathbb{R}^n$ shows that $\|e^{tA}\|_{\mathrm{op}}$ is bounded for $t \geq 0$, so $0$ is Lyapunov stable. Conversely, assume $0$ is Lyapunov stable. By the first step, there exists $K\geq 1$ such that \begin{align*} \|e^{tA_{\mathbb{C}}}\|_{\mathcal{L}(\mathbb{C}^n)} \leq K \end{align*} for every $t \geq 0$. Let $\lambda$ be an eigenvalue and let $v\neq 0$ be an eigenvector. Then \begin{align*} e^{tA_{\mathbb{C}}}v=e^{t\lambda}v. \end{align*} The bound gives \begin{align*} e^{t\operatorname{Re}(\lambda)}|v|_{\mathbb{C}^n}\leq K|v|_{\mathbb{C}^n} \end{align*} for all $t\geq 0$, and therefore $\operatorname{Re}(\lambda)\leq 0$. It remains to exclude nontrivial Jordan blocks on the imaginary axis. Suppose $\operatorname{Re}(\lambda)=0$ and that the Jordan block for $\lambda$ has size at least $2$. Then there exists a generalized eigenvector $w\in\mathbb{C}^n$ such that \begin{align*} (A_{\mathbb{C}}-\lambda I_n)w=v \end{align*} for some eigenvector $v\neq 0$ satisfying $(A_{\mathbb{C}}-\lambda I_n)v=0$. On the two-dimensional Jordan chain generated by $w$ and $v$, the exponential satisfies \begin{align*} e^{tA_{\mathbb{C}}}w=e^{t\lambda}(w+tv). \end{align*} Since $|e^{t\lambda}|=1$, the vector $w+tv$ has norm growing at least linearly up to a fixed additive constant: \begin{align*} |w+tv|_{\mathbb{C}^n}\geq t|v|_{\mathbb{C}^n}-|w|_{\mathbb{C}^n}. \end{align*} Thus $\sup_{t\geq 0}|e^{tA_{\mathbb{C}}}w|_{\mathbb{C}^n}=\infty$, contradicting boundedness of $e^{tA_{\mathbb{C}}}$. Hence every eigenvalue on the imaginary axis is semisimple. [guided] We now identify exactly when the family of operators $e^{tA}$ stays bounded. Lyapunov stability is weaker than attraction: solutions are allowed to rotate or remain constant, but they cannot grow. Assume first that all eigenvalues satisfy $\operatorname{Re}(\lambda)\leq 0$ and that every eigenvalue on the imaginary axis is semisimple. Work in the complex Jordan form \begin{align*} A_{\mathbb{C}} = SJS^{-1}. \end{align*} For a block $J_j=\lambda_j I_{m_j}+N_j$, the explicit formula is \begin{align*} e^{tJ_j}=e^{t\lambda_j}\sum_{k=0}^{m_j-1}\frac{t^kN_j^k}{k!}. \end{align*} If $\operatorname{Re}(\lambda_j)<0$, the exponential decay dominates the polynomial factor. More precisely, the polynomial \begin{align*} P_j(t):=\sum_{k=0}^{m_j-1}\frac{t^k\|N_j^k\|_{\mathcal{L}(\mathbb{C}^{m_j})}}{k!} \end{align*} satisfies \begin{align*} \|e^{tJ_j}\|_{\mathcal{L}(\mathbb{C}^{m_j})}\leq e^{t\operatorname{Re}(\lambda_j)}P_j(t). \end{align*} Because $\operatorname{Re}(\lambda_j)<0$, the function $t\mapsto e^{t\operatorname{Re}(\lambda_j)}P_j(t)$ is bounded on $[0,\infty)$. Let $B_j>0$ be an upper bound. If $\operatorname{Re}(\lambda_j)=0$, the semisimplicity hypothesis is used exactly here. It says that the Jordan block has size $1$, so $N_j=0$. Therefore \begin{align*} e^{tJ_j}=e^{t\lambda_j}I_1. \end{align*} Since $\operatorname{Re}(\lambda_j)=0$, we get \begin{align*} \|e^{tJ_j}\|_{\mathcal{L}(\mathbb{C})}=|e^{t\lambda_j}|=1. \end{align*} Thus every block is bounded, and because there are only finitely many blocks, $e^{tJ}$ is bounded. Multiplying by the fixed change-of-basis matrices gives \begin{align*} \|e^{tA_{\mathbb{C}}}\|_{\mathcal{L}(\mathbb{C}^n)} \leq \|S\|_{\mathcal{L}(\mathbb{C}^n)} \|S^{-1}\|_{\mathcal{L}(\mathbb{C}^n)} \|e^{tJ}\|_{\mathcal{L}(\mathbb{C}^n)}. \end{align*} Hence $e^{tA_{\mathbb{C}}}$ is bounded, and restricting to real initial data gives Lyapunov stability of $0$. Now assume $0$ is Lyapunov stable. From the first step, this means the operators $e^{tA_{\mathbb{C}}}$ are uniformly bounded: \begin{align*} \|e^{tA_{\mathbb{C}}}\|_{\mathcal{L}(\mathbb{C}^n)}\leq K \end{align*} for some $K\geq 1$ and all $t\geq 0$. First take an eigenvector $v\neq 0$ with eigenvalue $\lambda$. Then \begin{align*} e^{tA_{\mathbb{C}}}v=e^{t\lambda}v. \end{align*} The uniform bound gives \begin{align*} e^{t\operatorname{Re}(\lambda)}|v|_{\mathbb{C}^n}\leq K|v|_{\mathbb{C}^n}. \end{align*} After dividing by $|v|_{\mathbb{C}^n}>0$, we see that $e^{t\operatorname{Re}(\lambda)}\leq K$ for all $t\geq 0$. This forces $\operatorname{Re}(\lambda)\leq 0$. The remaining issue is the nilpotent part on the imaginary axis. Suppose $\operatorname{Re}(\lambda)=0$ and the Jordan block for $\lambda$ has size at least $2$. Then there is a Jordan chain of length $2$: an eigenvector $v\neq 0$ and a generalized eigenvector $w$ such that \begin{align*} (A_{\mathbb{C}}-\lambda I_n)v=0 \end{align*} and \begin{align*} (A_{\mathbb{C}}-\lambda I_n)w=v. \end{align*} On this chain, the nilpotent part contributes a linear term: \begin{align*} e^{tA_{\mathbb{C}}}w=e^{t\lambda}(w+tv). \end{align*} Because $\operatorname{Re}(\lambda)=0$, the scalar $e^{t\lambda}$ has modulus $1$. Hence \begin{align*} |e^{tA_{\mathbb{C}}}w|_{\mathbb{C}^n}=|w+tv|_{\mathbb{C}^n}. \end{align*} The [reverse triangle inequality](/theorems/2300) gives \begin{align*} |w+tv|_{\mathbb{C}^n}\geq t|v|_{\mathbb{C}^n}-|w|_{\mathbb{C}^n}. \end{align*} Since $v\neq 0$, the right-hand side tends to infinity as $t\to\infty$. This contradicts the assumed uniform boundedness of $e^{tA_{\mathbb{C}}}$. Therefore no imaginary-axis eigenvalue can have a nontrivial Jordan block, which is exactly semisimplicity. [/guided] [/step] [step:Combine the three classifications] The second and third steps prove that exponential stability is equivalent to $A$ being Hurwitz. The fourth step proves that asymptotic stability is also equivalent to $A$ being Hurwitz. The fifth step proves that Lyapunov stability is equivalent to the spectral condition $\operatorname{Re}(\lambda)\leq 0$ for every eigenvalue, together with semisimplicity of every eigenvalue satisfying $\operatorname{Re}(\lambda)=0$. These are precisely the three assertions of the theorem. [/step]